Homework #6

Question 1: Lab 3.6.2, p. 110-113

library(ISLR)
library(MASS)

Goal: predict medv using 13 predictors such as: * rm (avg. # rooms per house) * age (avg. age house) * lstat (% households w/ low socioeconomic status)

# fix(Boston) <- commented out because every load triggers an imported data window
names(Boston)

##  [1] "crim"    "zn"      "indus"   "chas"    "nox"     "rm"      "age"    
##  [8] "dis"     "rad"     "tax"     "ptratio" "black"   "lstat"   "medv"

Building the Linear Model

attach(Boston)

lm.fit = lm(medv~lstat)

#lm.fit
#summary(lm.fit)

Note: if we try to do the lm.fit() without attaching Boston, then we are referencing columns/variables that are NOT within the scope of R’s object global environment

• We can get around the above issue by adding the Boston$ before the column names, but this is just cumbersome, and it’s easier to just attach the dataset

Exploring the Linear Model

names(lm.fit)

##  [1] "coefficients"  "residuals"     "effects"       "rank"         
##  [5] "fitted.values" "assign"        "qr"            "df.residual"  
##  [9] "xlevels"       "call"          "terms"         "model"

coef(lm.fit)

## (Intercept)       lstat 
##  34.5538409  -0.9500494

#lm.fit$coefficients would work just as well, but it is not "as safe" --> trying to figure out why

#Now, building confidence interval:

confint(lm.fit)

##                 2.5 %     97.5 %
## (Intercept) 33.448457 35.6592247
## lstat       -1.026148 -0.8739505

confint(lm.fit,level = 0.99)

##                 0.5 %     99.5 %
## (Intercept) 33.099101 36.0085810
## lstat       -1.050199 -0.8498995

Utilizing predict

Using predict to construct confidence intervals:

predict(lm.fit, 
        data.frame(lstat = c(5, 10, 15)),
        interval ="confidence")

##        fit      lwr      upr
## 1 29.80359 29.00741 30.59978
## 2 25.05335 24.47413 25.63256
## 3 20.30310 19.73159 20.87461

Interpretation:

• the 95% confidence interval associated with 10 is (24.474, 25.633)

Using predict to construct prediction

predict(lm.fit, 
        data.frame(lstat = c(5, 10, 15)),
        interval ="prediction")

##        fit       lwr      upr
## 1 29.80359 17.565675 42.04151
## 2 25.05335 12.827626 37.27907
## 3 20.30310  8.077742 32.52846

Interpretation:

• the 95% prediction interval associated with 10 is (12.828, 37.280)

Plotting LSRL

library(ggplot2)
plot(lstat, medv)
#abline() plots a line on the previous plot
abline(lm.fit)

ggplot(Boston, aes(x = lstat, y = medv)) +
  geom_point() +
  geom_smooth(method = "lm", se = FALSE) 

Playing around with the different abline and plot() variations

plot(lstat,medv)
abline(lm.fit,lwd=3)
abline(lm.fit, lwd = 3, col = "red")

plot(lstat, medv, col = "red")

plot(lstat, medv, pch = 20)

plot(lstat, medv, pch = "+")

plot(1:20, 1:20, pch = 1:20)

Viewing lm() results in one window

par(mfrow = c(2,2))
plot(lm.fit)

Plotting the Residuals

plot(predict(lm.fit), residuals(lm.fit))

plot(predict(lm.fit), rstudent(lm.fit))

rstudent: calculates the studentized residuals which is: * quotient resulting from residual / estimate of sd

plot(hatvalues(lm.fit))

which.max(hatvalues(lm.fit))

## 375 
## 375

• On basis of residual, evidence of non-linearity
• Leverage: measure of how far away the independent variable values are from other observations

Question 2: Lab 3.6.3, p. 113 - 115

Multiple Linear Regression

lm.mlrfit = lm(medv~lstat + age, data = Boston)
summary(lm.mlrfit)

## 
## Call:
## lm(formula = medv ~ lstat + age, data = Boston)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -15.981  -3.978  -1.283   1.968  23.158 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 33.22276    0.73085  45.458  < 2e-16 ***
## lstat       -1.03207    0.04819 -21.416  < 2e-16 ***
## age          0.03454    0.01223   2.826  0.00491 ** 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 6.173 on 503 degrees of freedom
## Multiple R-squared:  0.5513, Adjusted R-squared:  0.5495 
## F-statistic:   309 on 2 and 503 DF,  p-value: < 2.2e-16

To avoid applying MLR to all 13 variables, we do the following: ~.

lm.mlrfit = lm(medv~., data=Boston)
summary(lm.mlrfit)

## 
## Call:
## lm(formula = medv ~ ., data = Boston)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -15.595  -2.730  -0.518   1.777  26.199 
## 
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  3.646e+01  5.103e+00   7.144 3.28e-12 ***
## crim        -1.080e-01  3.286e-02  -3.287 0.001087 ** 
## zn           4.642e-02  1.373e-02   3.382 0.000778 ***
## indus        2.056e-02  6.150e-02   0.334 0.738288    
## chas         2.687e+00  8.616e-01   3.118 0.001925 ** 
## nox         -1.777e+01  3.820e+00  -4.651 4.25e-06 ***
## rm           3.810e+00  4.179e-01   9.116  < 2e-16 ***
## age          6.922e-04  1.321e-02   0.052 0.958229    
## dis         -1.476e+00  1.995e-01  -7.398 6.01e-13 ***
## rad          3.060e-01  6.635e-02   4.613 5.07e-06 ***
## tax         -1.233e-02  3.760e-03  -3.280 0.001112 ** 
## ptratio     -9.527e-01  1.308e-01  -7.283 1.31e-12 ***
## black        9.312e-03  2.686e-03   3.467 0.000573 ***
## lstat       -5.248e-01  5.072e-02 -10.347  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 4.745 on 492 degrees of freedom
## Multiple R-squared:  0.7406, Adjusted R-squared:  0.7338 
## F-statistic: 108.1 on 13 and 492 DF,  p-value: < 2.2e-16

In order to access different parts of the summary, we can do:

#returns the R^2
summary(lm.mlrfit)$r.sq

## [1] 0.7406427

# returns the RSE
summary(lm.mlrfit)$sigma

## [1] 4.745298

Using the vif function: Variance Inflation Factors

library(car)
vif(lm.mlrfit)

##     crim       zn    indus     chas      nox       rm      age      dis 
## 1.792192 2.298758 3.991596 1.073995 4.393720 1.933744 3.100826 3.955945 
##      rad      tax  ptratio    black    lstat 
## 7.484496 9.008554 1.799084 1.348521 2.941491

** Note: ** VIF helps to determine multicollinearity

Accessing all but one variable

lm.mlrfit.noage = lm(medv~.-age, data = Boston)
summary(lm.mlrfit.noage)

## 
## Call:
## lm(formula = medv ~ . - age, data = Boston)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -15.6054  -2.7313  -0.5188   1.7601  26.2243 
## 
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  36.436927   5.080119   7.172 2.72e-12 ***
## crim         -0.108006   0.032832  -3.290 0.001075 ** 
## zn            0.046334   0.013613   3.404 0.000719 ***
## indus         0.020562   0.061433   0.335 0.737989    
## chas          2.689026   0.859598   3.128 0.001863 ** 
## nox         -17.713540   3.679308  -4.814 1.97e-06 ***
## rm            3.814394   0.408480   9.338  < 2e-16 ***
## dis          -1.478612   0.190611  -7.757 5.03e-14 ***
## rad           0.305786   0.066089   4.627 4.75e-06 ***
## tax          -0.012329   0.003755  -3.283 0.001099 ** 
## ptratio      -0.952211   0.130294  -7.308 1.10e-12 ***
## black         0.009321   0.002678   3.481 0.000544 ***
## lstat        -0.523852   0.047625 -10.999  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 4.74 on 493 degrees of freedom
## Multiple R-squared:  0.7406, Adjusted R-squared:  0.7343 
## F-statistic: 117.3 on 12 and 493 DF,  p-value: < 2.2e-16

Updating the linear models with variables

lm.mlrfit.withage = update(lm.mlrfit, ~.-age)

Question 3: Lab 3.6.4, p. 115

Interaction Terms

summary(lm(medv~lstat*age, data = Boston))

## 
## Call:
## lm(formula = medv ~ lstat * age, data = Boston)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -15.806  -4.045  -1.333   2.085  27.552 
## 
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 36.0885359  1.4698355  24.553  < 2e-16 ***
## lstat       -1.3921168  0.1674555  -8.313 8.78e-16 ***
## age         -0.0007209  0.0198792  -0.036   0.9711    
## lstat:age    0.0041560  0.0018518   2.244   0.0252 *  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 6.149 on 502 degrees of freedom
## Multiple R-squared:  0.5557, Adjusted R-squared:  0.5531 
## F-statistic: 209.3 on 3 and 502 DF,  p-value: < 2.2e-16

Note: The above is shorthand for lstat + age + lstat:age

Question 4: Lab 3.6.5, p. 115 - 117

Non-linear Transformations of Predictors

lm.fit.square = lm(medv~lstat + I(lstat^2))
summary(lm.fit.square)

## 
## Call:
## lm(formula = medv ~ lstat + I(lstat^2))
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -15.2834  -3.8313  -0.5295   2.3095  25.4148 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 42.862007   0.872084   49.15   <2e-16 ***
## lstat       -2.332821   0.123803  -18.84   <2e-16 ***
## I(lstat^2)   0.043547   0.003745   11.63   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 5.524 on 503 degrees of freedom
## Multiple R-squared:  0.6407, Adjusted R-squared:  0.6393 
## F-statistic: 448.5 on 2 and 503 DF,  p-value: < 2.2e-16

Note: With a near-zero p-value, this square model seems to be a better fit for the data

Comparing the models with anova()

anova(lm.fit, lm.fit.square)

## Analysis of Variance Table
## 
## Model 1: medv ~ lstat
## Model 2: medv ~ lstat + I(lstat^2)
##   Res.Df   RSS Df Sum of Sq     F    Pr(>F)    
## 1    504 19472                                 
## 2    503 15347  1    4125.1 135.2 < 2.2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

\(H_{0}\) : Two models fit the data equally well

\(H_{a}\) : Full model is superior (squared model)

Analysis:

• With F-stat of 135 and associated p-value of nearly 0, then it shows that the model with the linear and squared predictors is superior to the model with just the linear predictor

par(mfrow=c(2,2))
plot(lm.fit.square)

FOR NTH-TERM POLYNOMIALS, USE poly

lm.fit.fifth = lm(medv~poly(lstat, 5))
summary(lm.fit.fifth)

## 
## Call:
## lm(formula = medv ~ poly(lstat, 5))
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -13.5433  -3.1039  -0.7052   2.0844  27.1153 
## 
## Coefficients:
##                  Estimate Std. Error t value Pr(>|t|)    
## (Intercept)       22.5328     0.2318  97.197  < 2e-16 ***
## poly(lstat, 5)1 -152.4595     5.2148 -29.236  < 2e-16 ***
## poly(lstat, 5)2   64.2272     5.2148  12.316  < 2e-16 ***
## poly(lstat, 5)3  -27.0511     5.2148  -5.187 3.10e-07 ***
## poly(lstat, 5)4   25.4517     5.2148   4.881 1.42e-06 ***
## poly(lstat, 5)5  -19.2524     5.2148  -3.692 0.000247 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 5.215 on 500 degrees of freedom
## Multiple R-squared:  0.6817, Adjusted R-squared:  0.6785 
## F-statistic: 214.2 on 5 and 500 DF,  p-value: < 2.2e-16

• This shows that polynomials up to the 5th power are significant in the regression fit

lm.fit.sixth = lm(medv~poly(lstat, 6))
summary(lm.fit.sixth)

## 
## Call:
## lm(formula = medv ~ poly(lstat, 6))
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -14.7317  -3.1571  -0.6941   2.0756  26.8994 
## 
## Coefficients:
##                  Estimate Std. Error t value Pr(>|t|)    
## (Intercept)       22.5328     0.2317  97.252  < 2e-16 ***
## poly(lstat, 6)1 -152.4595     5.2119 -29.252  < 2e-16 ***
## poly(lstat, 6)2   64.2272     5.2119  12.323  < 2e-16 ***
## poly(lstat, 6)3  -27.0511     5.2119  -5.190 3.06e-07 ***
## poly(lstat, 6)4   25.4517     5.2119   4.883 1.41e-06 ***
## poly(lstat, 6)5  -19.2524     5.2119  -3.694 0.000245 ***
## poly(lstat, 6)6    6.5088     5.2119   1.249 0.212313    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 5.212 on 499 degrees of freedom
## Multiple R-squared:  0.6827, Adjusted R-squared:  0.6789 
## F-statistic: 178.9 on 6 and 499 DF,  p-value: < 2.2e-16

• But this analysis shows that any terms beyond the fifth power are NOT signficant predictors

USING LOG PREDICTORS

 summary(lm(medv~log(rm), data = Boston))

## 
## Call:
## lm(formula = medv ~ log(rm), data = Boston)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -19.487  -2.875  -0.104   2.837  39.816 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  -76.488      5.028  -15.21   <2e-16 ***
## log(rm)       54.055      2.739   19.73   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 6.915 on 504 degrees of freedom
## Multiple R-squared:  0.4358, Adjusted R-squared:  0.4347 
## F-statistic: 389.3 on 1 and 504 DF,  p-value: < 2.2e-16

Question 5: Lab 3.6.6, p. 117 - 119

Qualitative Predictors

#fix(Carseats) <- commented out because every load triggers an imported data window
names(Carseats)

##  [1] "Sales"       "CompPrice"   "Income"      "Advertising" "Population" 
##  [6] "Price"       "ShelveLoc"   "Age"         "Education"   "Urban"      
## [11] "US"

lm.car.fit = lm(Sales ~. + Income:Advertising + Price:Age, data = Carseats)
summary(lm.car.fit)

## 
## Call:
## lm(formula = Sales ~ . + Income:Advertising + Price:Age, data = Carseats)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -2.9208 -0.7503  0.0177  0.6754  3.3413 
## 
## Coefficients:
##                      Estimate Std. Error t value Pr(>|t|)    
## (Intercept)         6.5755654  1.0087470   6.519 2.22e-10 ***
## CompPrice           0.0929371  0.0041183  22.567  < 2e-16 ***
## Income              0.0108940  0.0026044   4.183 3.57e-05 ***
## Advertising         0.0702462  0.0226091   3.107 0.002030 ** 
## Population          0.0001592  0.0003679   0.433 0.665330    
## Price              -0.1008064  0.0074399 -13.549  < 2e-16 ***
## ShelveLocGood       4.8486762  0.1528378  31.724  < 2e-16 ***
## ShelveLocMedium     1.9532620  0.1257682  15.531  < 2e-16 ***
## Age                -0.0579466  0.0159506  -3.633 0.000318 ***
## Education          -0.0208525  0.0196131  -1.063 0.288361    
## UrbanYes            0.1401597  0.1124019   1.247 0.213171    
## USYes              -0.1575571  0.1489234  -1.058 0.290729    
## Income:Advertising  0.0007510  0.0002784   2.698 0.007290 ** 
## Price:Age           0.0001068  0.0001333   0.801 0.423812    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.011 on 386 degrees of freedom
## Multiple R-squared:  0.8761, Adjusted R-squared:  0.8719 
## F-statistic:   210 on 13 and 386 DF,  p-value: < 2.2e-16

Viewing dummy variables generated by R

attach(Carseats)
contrasts(ShelveLoc)

##        Good Medium
## Bad       0      0
## Good      1      0
## Medium    0      1

Interpretation:

• ShelveLocGood: indicates positive output in good shelving location associated with high sales

• ShelveLocMedium: indicates positive output in medium shelving location associated with higher sales than bad shelving location but lower sales than good shelving location

Question 6: Exercise 8, p. 121 - 122

Exercise 8

importing the dataset:

#fix(Auto)<-commented out because every load triggers an imported data window
names(Auto)

## [1] "mpg"          "cylinders"    "displacement" "horsepower"  
## [5] "weight"       "acceleration" "year"         "origin"      
## [9] "name"

1. Using lm() to perform simple linear regression with mpg as response and horsepower as predictor

lm.auto.fit = lm(mpg~horsepower, data = Auto)
summary(lm.auto.fit)

## 
## Call:
## lm(formula = mpg ~ horsepower, data = Auto)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -13.5710  -3.2592  -0.3435   2.7630  16.9240 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 39.935861   0.717499   55.66   <2e-16 ***
## horsepower  -0.157845   0.006446  -24.49   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 4.906 on 390 degrees of freedom
## Multiple R-squared:  0.6059, Adjusted R-squared:  0.6049 
## F-statistic: 599.7 on 1 and 390 DF,  p-value: < 2.2e-16

plot(mpg~horsepower, data = Auto)

print("Predicted Value @ horsepower = 98")

## [1] "Predicted Value @ horsepower = 98"

predict(lm.auto.fit, 
        data.frame(horsepower =98))

##        1 
## 24.46708

print("Confidence Interval:")

## [1] "Confidence Interval:"

predict(lm.auto.fit, 
        data.frame(horsepower = 98),
        interval ="confidence",
        level =0.95)

##        fit      lwr      upr
## 1 24.46708 23.97308 24.96108

print("Prediction Interval: ")

## [1] "Prediction Interval: "

predict(lm.auto.fit, 
        data.frame(horsepower = 98),
        interval ="prediction", 
        level =0.95)

##        fit     lwr      upr
## 1 24.46708 14.8094 34.12476

• i, ii, & iii: There exists a strong, negative relationship between the predictor and the response

• iv:

  * At a `horsepower` of 98, the predicted `mpg` is 24.67 
  
  * The associated 95% **confidence interval** is (23.973, 24.961)
  
  * The associated 95% **prediction interval** is (14.809, 34.125)

2. Plotting

attach(Auto)

## The following object is masked from package:ggplot2:
## 
##     mpg

plot(horsepower, mpg) + 
  title("Plot of Horsepower vs. mpg")

## numeric(0)

abline(lm.auto.fit)

##ggplot 
ggplot(Auto, aes(x=horsepower, y = mpg)) +
  geom_point() +
  geom_smooth(method = "lm", se = FALSE) +
  ggtitle("Plot of Horsepower vs. mpg")

3. Plotting diagnostic plots of LSR

par(mfrow=c(2,2))
plot(lm.auto.fit)

Comments:

• There exists normality

• There exists Heteroscedasticity

• Not too much leverage

• Approximately parametric (definitively not absolutely linear)

Question 7: Exercise 9, p. 121

1. Scatterplot matrix with all of the variables

attach(Auto)

## The following objects are masked from Auto (pos = 3):
## 
##     acceleration, cylinders, displacement, horsepower, mpg, name,
##     origin, weight, year

## The following object is masked from package:ggplot2:
## 
##     mpg

pairs(mpg~.,
      data=Auto, 
      main="Simple Scatterplot Matrix")

#corplot

2. Matrix of Correlations b/w variables using cor(). Exclude name

cor(y= Auto$mpg, x = Auto[,2:8])

##                    [,1]
## cylinders    -0.7776175
## displacement -0.8051269
## horsepower   -0.7784268
## weight       -0.8322442
## acceleration  0.4233285
## year          0.5805410
## origin        0.5652088

3. Multiple Linear Regression

lm.auto.mlr = lm(mpg~.-name, data = Auto)
summary(lm.auto.mlr)

## 
## Call:
## lm(formula = mpg ~ . - name, data = Auto)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -9.5903 -2.1565 -0.1169  1.8690 13.0604 
## 
## Coefficients:
##                Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  -17.218435   4.644294  -3.707  0.00024 ***
## cylinders     -0.493376   0.323282  -1.526  0.12780    
## displacement   0.019896   0.007515   2.647  0.00844 ** 
## horsepower    -0.016951   0.013787  -1.230  0.21963    
## weight        -0.006474   0.000652  -9.929  < 2e-16 ***
## acceleration   0.080576   0.098845   0.815  0.41548    
## year           0.750773   0.050973  14.729  < 2e-16 ***
## origin         1.426141   0.278136   5.127 4.67e-07 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 3.328 on 384 degrees of freedom
## Multiple R-squared:  0.8215, Adjusted R-squared:  0.8182 
## F-statistic: 252.4 on 7 and 384 DF,  p-value: < 2.2e-16

coef(lm.auto.mlr)

##   (Intercept)     cylinders  displacement    horsepower        weight 
## -17.218434622  -0.493376319   0.019895644  -0.016951144  -0.006474043 
##  acceleration          year        origin 
##   0.080575838   0.750772678   1.426140495

• i, ii: There exists a relationship between some of the predictors and the response. The following predictors have a statistically significant relationship to the response:
  • displacement
  • weight
  • year
  • origin
• iii: The coefficient of year (0.751) suggests that there exists a strong positive relationship between year and mpg. Furthermore, this also suggests that for every additional year, mpg on a car increases by 0.751

4. Plotting the Linear Regression Fit

par(mfrow=c(2,2))
plot(lm.auto.mlr)

• Only one “Large” outlier –> the data point marked “14” indicates incredibly high leverage

*Approximately homoscedatic

• Approximately linear

• Definitely linear

e)Checking Interactions

attach(Auto)

## The following objects are masked from Auto (pos = 3):
## 
##     acceleration, cylinders, displacement, horsepower, mpg, name,
##     origin, weight, year

## The following objects are masked from Auto (pos = 4):
## 
##     acceleration, cylinders, displacement, horsepower, mpg, name,
##     origin, weight, year

## The following object is masked from package:ggplot2:
## 
##     mpg

lm(mpg~cylinders:displacement)

## 
## Call:
## lm(formula = mpg ~ cylinders:displacement)
## 
## Coefficients:
##            (Intercept)  cylinders:displacement  
##              30.989620               -0.006118

lm(mpg~cylinders:horsepower)

## 
## Call:
## lm(formula = mpg ~ cylinders:horsepower)
## 
## Coefficients:
##          (Intercept)  cylinders:horsepower  
##             32.49813              -0.01444

lm(mpg~cylinders:weight)

## 
## Call:
## lm(formula = mpg ~ cylinders:weight)
## 
## Coefficients:
##      (Intercept)  cylinders:weight  
##       34.2947992        -0.0006168

lm(mpg~cylinders:acceleration)

## 
## Call:
## lm(formula = mpg ~ cylinders:acceleration)
## 
## Coefficients:
##            (Intercept)  cylinders:acceleration  
##                40.9279                 -0.2115

lm(mpg~cylinders:year)

## 
## Call:
## lm(formula = mpg ~ cylinders:year)
## 
## Coefficients:
##    (Intercept)  cylinders:year  
##       42.48053        -0.04602

lm(mpg~cylinders:origin)

## 
## Call:
## lm(formula = mpg ~ cylinders:origin)
## 
## Coefficients:
##      (Intercept)  cylinders:origin  
##           19.664             0.482

lm(mpg~displacement:horsepower)

## 
## Call:
## lm(formula = mpg ~ displacement:horsepower)
## 
## Coefficients:
##             (Intercept)  displacement:horsepower  
##              29.8890103               -0.0002694

lm(mpg~displacement:weight)

## 
## Call:
## lm(formula = mpg ~ displacement:weight)
## 
## Coefficients:
##         (Intercept)  displacement:weight  
##           3.126e+01           -1.182e-05

lm(mpg~displacement:acceleration)

## 
## Call:
## lm(formula = mpg ~ displacement:acceleration)
## 
## Coefficients:
##               (Intercept)  displacement:acceleration  
##                 36.933531                  -0.004708

lm(mpg~displacement:year)

## 
## Call:
## lm(formula = mpg ~ displacement:year)
## 
## Coefficients:
##       (Intercept)  displacement:year  
##          35.29577           -0.00081

lm(mpg~displacement:origin)

## 
## Call:
## lm(formula = mpg ~ displacement:origin)
## 
## Coefficients:
##         (Intercept)  displacement:origin  
##            33.44016             -0.03922

lm(mpg~horsepower:weight)

## 
## Call:
## lm(formula = mpg ~ horsepower:weight)
## 
## Coefficients:
##       (Intercept)  horsepower:weight  
##         3.292e+01         -2.791e-05

lm(mpg~horsepower:acceleration)

## 
## Call:
## lm(formula = mpg ~ horsepower:acceleration)
## 
## Coefficients:
##             (Intercept)  horsepower:acceleration  
##                47.72990                 -0.01566

lm(mpg~horsepower:year)

## 
## Call:
## lm(formula = mpg ~ horsepower:year)
## 
## Coefficients:
##     (Intercept)  horsepower:year  
##       40.451416        -0.002158

lm(mpg~horsepower:origin)

## 
## Call:
## lm(formula = mpg ~ horsepower:origin)
## 
## Coefficients:
##       (Intercept)  horsepower:origin  
##         23.232932           0.001414

lm(mpg~weight:acceleration)

## 
## Call:
## lm(formula = mpg ~ weight:acceleration)
## 
## Coefficients:
##         (Intercept)  weight:acceleration  
##          40.5317687           -0.0003772

lm(mpg~weight:year)

## 
## Call:
## lm(formula = mpg ~ weight:year)
## 
## Coefficients:
## (Intercept)  weight:year  
##   4.571e+01   -9.882e-05

lm(mpg~weight:origin)

## 
## Call:
## lm(formula = mpg ~ weight:origin)
## 
## Coefficients:
##   (Intercept)  weight:origin  
##     2.030e+01      7.315e-04

lm(mpg~acceleration:year)

## 
## Call:
## lm(formula = mpg ~ acceleration:year)
## 
## Coefficients:
##       (Intercept)  acceleration:year  
##           2.56177            0.01764

lm(mpg~acceleration:origin)

## 
## Call:
## lm(formula = mpg ~ acceleration:origin)
## 
## Coefficients:
##         (Intercept)  acceleration:origin  
##             14.9387               0.3407

f)Varying Transformations

lm.auto.sq = lm(mpg+Auto[,2:8]+ I(Auto[,2:8])^2, data = Auto)
summary(lm.auto.sq)

## 
## Call:
## lm(formula = mpg + Auto[, 2:8] + I(Auto[, 2:8])^2, data = Auto)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -1.39763 -0.37385  0.00738  0.21238  1.49877 
## 
## Coefficients:
##                Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   2.5424031  0.7206057   3.528 0.000469 ***
## displacement  0.0154275  0.0008862  17.408  < 2e-16 ***
## horsepower   -0.0052229  0.0021571  -2.421 0.015929 *  
## weight        0.0002567  0.0001020   2.518 0.012208 *  
## acceleration -0.0154412  0.0155628  -0.992 0.321731    
## year         -0.0024005  0.0080349  -0.299 0.765289    
## origin        0.0848696  0.0436337   1.945 0.052497 .  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.5246 on 385 degrees of freedom
## Multiple R-squared:  0.9069, Adjusted R-squared:  0.9054 
## F-statistic: 624.8 on 6 and 385 DF,  p-value: < 2.2e-16

lm.auto.3 = lm(mpg+Auto[,2:8]+I(Auto[,2:8])^3, data = Auto)
summary(lm.auto.3)

## 
## Call:
## lm(formula = mpg + Auto[, 2:8] + I(Auto[, 2:8])^3, data = Auto)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -1.39763 -0.37385  0.00738  0.21238  1.49877 
## 
## Coefficients:
##                Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   2.5424031  0.7206057   3.528 0.000469 ***
## displacement  0.0154275  0.0008862  17.408  < 2e-16 ***
## horsepower   -0.0052229  0.0021571  -2.421 0.015929 *  
## weight        0.0002567  0.0001020   2.518 0.012208 *  
## acceleration -0.0154412  0.0155628  -0.992 0.321731    
## year         -0.0024005  0.0080349  -0.299 0.765289    
## origin        0.0848696  0.0436337   1.945 0.052497 .  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.5246 on 385 degrees of freedom
## Multiple R-squared:  0.9069, Adjusted R-squared:  0.9054 
## F-statistic: 624.8 on 6 and 385 DF,  p-value: < 2.2e-16

lm.auto.4= lm(mpg+Auto[,2:8]+I(Auto[,2:8])^4, data = Auto)
summary(lm.auto.4)

## 
## Call:
## lm(formula = mpg + Auto[, 2:8] + I(Auto[, 2:8])^4, data = Auto)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -1.39763 -0.37385  0.00738  0.21238  1.49877 
## 
## Coefficients:
##                Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   2.5424031  0.7206057   3.528 0.000469 ***
## displacement  0.0154275  0.0008862  17.408  < 2e-16 ***
## horsepower   -0.0052229  0.0021571  -2.421 0.015929 *  
## weight        0.0002567  0.0001020   2.518 0.012208 *  
## acceleration -0.0154412  0.0155628  -0.992 0.321731    
## year         -0.0024005  0.0080349  -0.299 0.765289    
## origin        0.0848696  0.0436337   1.945 0.052497 .  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.5246 on 385 degrees of freedom
## Multiple R-squared:  0.9069, Adjusted R-squared:  0.9054 
## F-statistic: 624.8 on 6 and 385 DF,  p-value: < 2.2e-16

lm.auto.log= lm(mpg+Auto[,2:8]+log(Auto[,2:8]), data = Auto)
summary(lm.auto.log)

## 
## Call:
## lm(formula = mpg + Auto[, 2:8] + log(Auto[, 2:8]), data = Auto)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -1.39763 -0.37385  0.00738  0.21238  1.49877 
## 
## Coefficients:
##                Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   2.5424031  0.7206057   3.528 0.000469 ***
## displacement  0.0154275  0.0008862  17.408  < 2e-16 ***
## horsepower   -0.0052229  0.0021571  -2.421 0.015929 *  
## weight        0.0002567  0.0001020   2.518 0.012208 *  
## acceleration -0.0154412  0.0155628  -0.992 0.321731    
## year         -0.0024005  0.0080349  -0.299 0.765289    
## origin        0.0848696  0.0436337   1.945 0.052497 .  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.5246 on 385 degrees of freedom
## Multiple R-squared:  0.9069, Adjusted R-squared:  0.9054 
## F-statistic: 624.8 on 6 and 385 DF,  p-value: < 2.2e-16

I had difficulty doing computing squares/logs with MLR, so I’ll follow up in class with questions